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How Switch Mode Power Supply (SMPS) Circuits Work

Last Updated on by 58 Comments

SMPS is the acronym of the word Switch Mode Power Supply. The name clearly suggests that the concept has something or entirely to do with pulses or switching of the employed devices. I have explained how SMPS adapters work for converting mains voltage to a lower DC voltage.

Table of Contents

Advantage of SMPS Topology

In SMPS adapters the idea is to switch the mains input voltage into the primary winding of a transformer so that a lower value DC voltage may be obtained at the secondary winding of the transformer.

However the question is, the same can be done with an ordinary transformer, so what is the need of such complicated configuration when the functioning can be simply implemented though ordinary transformers?

Well, the concept was developed precisely for eliminating the use of heavy and bulky transformers with much efficient versions of SMPS power supply circuits.

Though the principle of operation is quite the similar, the results are hugely different.

Our mains voltage is also a pulsating voltage or an AC which is normally fed into the ordinary transformer for the required conversions, but we cannot make the transformer smaller in size even with current as low as 500 mA.

The reason behind this is the very low frequency involved with our AC mains inputs.
At 50 Hz or 60 Hz, the value is tremendously low for implementing them into high DC currents outputs using smaller transformers.

This is because as the frequency decrease, the eddy current losses with the transformer magnetization increases, which results in huge lose of current through heat and subsequently the whole process becomes very inefficient.

To compensate the above loss, relatively larger transformer cores are involved with relevant degree of wire thickness, making the entire unit heavy and cumbersome.

A switch mode power supply circuit tackles this issue very cleverly.

If lower frequency increases eddy current losses, means an increase in the frequency would do just the opposite.

Meaning if the frequency is increased, the transformer could be made much smaller yet would provide higher current at their outputs.

10WSMPSCircuitUsingSTMicroelectronicsVIPer22

That's exactly what we do with an SMPS circuit. So I have explained the functioning with the following points:

How SMPS adapters work

In a switch mode power supply circuit diagram, the input AC is first rectified and filtered to produce relevant magnitude of DC.

The above DC is applied to an oscillator configuration comprising a high voltage transistor or a mosfet, rigged to a well dimensioned small ferrite transformer primary winding.

The circuit becomes a self oscillating type of configuration which starts oscillating at some predetermined frequency set by other passive components like capacitors and resistors.

The frequency is usually above 50 Khz.

This frequency induces an equivalent voltage and current at the secondary winding of the transformer, determined by the number of turns and the SWG of the wire.

Due the involvement of high frequencies, eddy current losses become negligibly small and high current DC output becomes derivable through smaller ferrite cored transformers and relatively thinner wire winding.

However the secondary voltage will also be at the primary frequency, therefore it is once again rectified and filtered using a fast recovery diode and a high value capacitor.

The result at the output is a perfectly filtered low DC, which can be used effectively for operating any electronic circuit.

In modern versions of SMPS, hi-end ICs are employed instead of transistors at the input.
The ICs are equipped with a built in high voltage mosfet for sustaining high frequency oscillations and many other protection features.

What Built-in Protections do SMPS Have

These ICs have adequate built in protection circuitry like avalanche protection, over heat protection and output over voltage protection and also a burst mode feature.

Avalanche protection ensures that the IC does not get damaged during power switch ON current in rush.

The over heat protection ensures that the IC is automatically shuts off if the transformer is not wound correctly and draws more current from the IC making it dangerously hot.

The burst mode is an interesting feature included with the modern SMPS units.

Here, the output DC is fed back to a sensing input of the IC. If due to some reason, normally due to wrong secondary winding or selection of resistors the output voltage rises above a certain predetermined value, the IC shuts off the input switching and skips the switching into intermittent bursts.

This helps to control the voltage at the output and also the current at the output.

The feature also ensures that if the the output voltage is adjusted to some high point and the output is not loaded, the IC switches to burst mode making sure that the unit is operated intermittently until the output gets adequately loaded, this saves power of the unit when in standby conditions or when the output is not operative.

The feedback from the output section to the IC is implemented via an opto-coupler so that the output remains well aloof from the input high voltage mains AC, avoiding dangerous shocks.

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Comments

  1. Dear sir, how do I use this formula to calculate link capacitor?

    Cin= 2×Pin ×(0.25+1/pi ×ArcSin(Vbulk(min)/√² ×Vin(min))) / (2×Vin(min)² – Vbulk(min)²)× fline(min)

    Where:
    VIN Input Voltage 85 –115/230 — 265 VRMS
    fLINE Line Frequency 47 –50/60 — 63 Hz
    VOUT Output Voltage IOUT(min) ≤ IOUT ≤ IOUT(max) 11.75 — 12 –12.25 V
    VRIPPLE Output Ripple Voltage IOUT(min) ≤ IOUT ≤ IOUT(max) 100 mVpp
    IOUT Output Current 0 — 4 A
    η Efficiency 85%

    What I am trying is below:

    First, the denominator:
    2×Vin(min)² – Vbulk(min)² × fline(min)
    (2 × 85² – 75² ) × 47
    = 414775

    And Second, the numerator:
    2 × Pin ×(0.25+ 1/pi × ArcSin(Vbulk(min)/√2 ×Vin(min)))
    2 × Pin ×(0.25+ 1/pi × ArcSin(75/ √2×85))
    2 × Pin ×(0.25+ 1/pi × ArcSin(75/ 120.208))
    2 × Pin ×(0.25+ 1/pi × ArcSin(0.6239))
    2 ×Pin ×(0.25+ 1/pi × 38.6029)
    2 ×Pin ×(12.5377)

    But the resulr is too huge for Power in watt:

       126uF =  2× Pin × 12.5377 / 414775

    My questions:

    1. Is the denominator correct?
    2. Is the numerator correct?
    3. Why wattage is so hight if the above calculation are all correct

    Thank you sir

    Kamus

    Reply

    • I am sorry Kamus, it might take a lot of time to investigate your calculations, and due to lack of time it may be difficult for me to figure it out.

      I can only help with general circuit ideas.

      Reply

  3. Hello!Atx power supplies have a switch for 110v-220v selection.How does this switch work?
    Atx power supply has two large filter capacitors in the input section. Usually 470 microfarad is 200V.Why is the capacitor not one?
    470 microfarad 200V equivalent to two capacitors 1000 microfarad 400V one can not be used?

    Reply

    • The switch must be used depending on the input voltage specification. For 220V inputs, slide it towards the 220V side and for 120V supply sockets slide it on the 120V side. I have not checked the capacitor and connection details, so I cannot advise on that

      Reply

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